Question

2x^3y + 18xy - 10x^2y - 90y

Part A: Rewrite the expression so that the GCF is factored completely. (3 points)

Answer 1

First off, look carefully and decide whether there is any one variable that is common to all four terms. If yes, identify that variable / factor and factor it out of \[2x^3y + 18xy - 10x^2y - 90y\]

Answer 2

Next, look at the pairs 2, 18 and -10, -90. Any common factor(s)?

Answer 3

Wouldn't 2 be the only one since thats the smallest divisible number?

Answer 4

@mathmale

Answer 5

Somebody please help me this is a timed assessment and I have to turn it in A.S.A.P.

Answer 6

@mlb004

Answer 7

Pleeeeeeeaaaaaseeee help me

Answer 8

@rational can you please help

Answer 9

Certainly 2 can be factored out of all four terms. What other quantity can also be factored out?

Answer 10

1?

Answer 11

Factoring out the '2' results in
\[2x^3y + 18xy - 10x^2y - 90y=2(x^3y+9xy-5x^2y-45y).\] Double check that and let me know whether you're comfortable with this factoring out of the '2.'

Answer 12

'1' certainly is a factor, but it's a trivial factor, and you need not worry about it here.

Answer 13

http://openstudy.com/study#/updates/531f8d1fe4b040e0bc630008

Answer 14

Yes so basically everything would just be divided by 2?

Answer 15

Your original expression has four terms. yes, divide each of these four terms by 2, writing the resulting factors as I showed you, above.

Answer 16

Now please look again at \[2(x^3y+9xy-5x^2y-45y)\]

Answer 17

Specifically, look at the four terms inside the set of parentheses.

Answer 18

These four terms have one factor in common. What is that factor?

Answer 19

x^2y?

Answer 20

Ask yourself: is your x^2y a factor of the last term, -45y?

Answer 21

Oh it would be y

Answer 22

Right. So, y is common to all four terms within ( ) and can thus be factored out.

Do that now. Your result should look like 2y( ... ??? ... )

Answer 23

I don't know how to do that. I tried to access my course for help on that end but it is locked during the assessment.

Answer 24

You have:\[2(x^3y+9xy-5x^2y-45y)\]

Answer 25

Since multiplying this by (y/y) or 1 does not change the value of the expression, factor by multiplying that 2 by y and then dividing every one of the four terms within parentheses by y. Do that now, please.

Answer 26

Okay I can do that. Hang on one sec.

Answer 27

\[(x^2y+9xy-5x^2y-45y)\div2y\]

Answer 28

Is that the proper setup?

Answer 29

\[2y(\frac{ x^3y }{ y }+9\frac{ xy }{ y }-5\frac{ x^2y }{ y }-45\frac{ y }{ y })\] Please compare our results. Notice that 2y MULTIPLIES what's within parentheses; we divide each of the four terms within parentheses by y (not by 2y).

Answer 30

\[2y(x^2+9x-5x^2-45)\]

Answer 31

Very nice. Now, do you see that the last two terms inside the parentheses have a common factor? What is it?

Answer 32

5

Answer 33

Yes. Please take your last expression and factor 5 out of the last two terms.

Answer 34

\[2y(x^2+9x-x^2-9)\]

Answer 35

\[2y(x^2+9x-5x^2-45) ~becomes~2y(x^2+9x-5[(5/5)x^2-9)]\]

Answer 36

You are to factor out the 5, but that doesn't mean you can throw it out entirely. Think of factoring as re-arranging terms. The 5 from both of the last 2 terms becomes a multiplier.
5x^2 becomes 5(x^2), and -45 becomes 5(-9).

Answer 37

iN EACH case I've factored out the 5, but have not discarded the 5.

Answer 38

*brain explodes*

Answer 39

Sorry!! What could I do at this point that would most help you?

Answer 40

To answer my own question: You have:\[2y(x^2+9x-x^2-9)\] which lacks that '5' you were to factor out. I'll put back in that '5' with careful use of parentheses.

Answer 41

2y(x^2+9x-5[x^2-9)])

Answer 42

I'm not really sure. The fact that this is still the first question kinda scares me. Haha. I've gotten this information. How can I organize it to fit the needs of my question?

An expression is shown below:

2x3y + 18xy - 10x2y - 90y

Part A: Rewrite the expression so that the GCF is factored completely. (3 points)

Part B: Rewrite the expression completely factored. Show the steps of your work. (4 points)

Part C: If the two middle terms were switched so that the expression became 2x^3y - 10x^2y + 18xy - 90y, would the factored expression no longer be equivalent to your answer in part B? Explain your reasoning. (3 points)

Simplifying
2x^3y + 18xy + -10x^2y + -90y

Reorder the terms:
18xy + -10x^2y + 2x^3y + -90y

Factor out the Greatest Common Factor (GCF), '2y'.
2y(9x + -5x^2 + x^3 + -45)

Final result:
2y(9x + -5x^2 + x^3 + -45)

Answer 43

Just a moment, please.

Answer 44

Part A: Factor out the GCF:\[2(x^3y+9xy-5x^2y-45y)\rightarrow 2y(x^3+9x-5x^2-45)\]

Answer 45

Are y ou OK with this? If not, ask questions.

Answer 46

No I completely understand that

Answer 47

Next step: look at the first 2 terms within parentheses and factor out the common factor.

Answer 48

Thats where you lose me

Answer 49

2y(x^3+9x-5x^2-45) becomes 2y(x(x^2-9) - 5(x^2-9) )

Answer 50

Inside parentheses we had x^3+9x as the first two terms, right?

Answer 51

Yes

Answer 52

And that translates into 2y ( [x^2-9](x-5) ).

Answer 53

Why? Because x^2-9 is a multiplier of both the x and the -5. OK or not OK?

Answer 54

I understand

Answer 55

Brb. Coffee and bathroom break. Lol

Answer 56

Cool. So, then, we have 2y (x^2-9) (x-5).

Answer 57

Note: This approach is called FACTORING BY GROUPING.

Answer 58

B-C: I need to move on.
We've used factoring by grouping to get where we are now.
Note that x^2-9 is the difference of two squares, and as such can be factored further. What are the factors of \[x^2?~ of 9?\]

Answer 59

Sorry i'm back. Wouldnt x^2 be the only factor?

Answer 60

No. Both x^2 and 9 are perfect squares. What is the square root of each?

Answer 61

It is well worth learning how to factor the difference of two squares; you will see this kind of situation often.

Answer 62

Your final expression, involving nothing that is further factorable, should look like\[2y (~~~~)(~~~~)(x-5)\]

Answer 63

Look up "difference of two squares" on the Internet if necessary. You're likely to find examples there.